In how many ways can we colour 25 white balls with 6 colours such that

a) each colour is used at least once and no ball remains uncoloured
b) all balls are coloured; I thought it would be $6^{25}$
c) no ball uncoloured, exactly 3 colours used. $\frac{6^{25}}{6\choose 3}$?

  • $\begingroup$ What are your thoughts on (a)? $\endgroup$
    – BallBoy
    Feb 8, 2018 at 13:59
  • 2
    $\begingroup$ If you have distinguishable balls (e.g. if they are numbered or otherwise marked to tell them apart), then your idea for part (b) is correct. But I suspect the balls were meant to be all alike before coloring, so that counting the ways to color them means just different outcomes of six color combinations with repetitions. $\endgroup$
    – hardmath
    Feb 8, 2018 at 14:04

2 Answers 2


I'll wait for your thoughts to respond to (a).

(b) is good: since you have 6 choices for each ball, and 25 balls, the total number of ways is $6 \cdot 6 \cdots 6$ 25 times, or $6^{25}$.

For (c), if you knew a specific set of 3 colors were being used, and it were "up to 3 colors," not "exactly," it would be $3^{25}$, by the same logic as (b). Since we need exactly 3 colors, we need to apply the Principle of Inclusion-Exclusion (are you familiar with this?) to get $3^{25} - 3\cdot2^{25} + 3\cdot1^{25}$. Finally, since there are $\binom63$ choices of sets of 3 colors, the answer should be $\binom63(3^{25} - 3\cdot2^{25} + 3\cdot1^{25})$.


(a) $x_1+x_2+x_3+x_4+x_5+x_6=25$

None of the x values are zero and using STARS and BARS method for positive integer

$\binom{25-1}{6-1} = \binom{24}{5} = \frac{24!}{5!19!} = 42504$

(b) $x_1+x_2+x_3+x_4+x_5+x_6=25$

Using STARS and BARS method for non negative solution

$\binom{25+6-1}{6-1} = \binom{30}{5} = \frac{30!}{5!25!} = 142506$

(c)three colours can be chosen by$\binom{6}{3}$


Using STARS and BARS method for non negative solution

$\binom{25+3-1}{3-1} = \binom{27}{2} = \frac{27!}{2!25!} = 351$



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